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A335018
Number of triples (d1,d2,d3) where each element is a divisor of m and d1 + d2 + d3 <= m where m is least odd integer of each prime signature.
0
0, 1, 8, 27, 27, 125, 64, 343, 343, 512, 125, 1331, 729, 1331, 216, 3375, 3375, 1331, 4913, 2744, 343, 6859, 3375, 12167, 2197, 12167, 4913, 512, 12167, 6859, 29791, 3375, 17576, 24389, 29791, 42875, 8000, 729, 29791, 19683, 12167, 59319, 4913, 42875, 42875, 103823, 13824
OFFSET
1,3
COMMENTS
All terms are cubes. Proof: Let d_k be the k-th divisor of some odd m > 1 and t be the number of divisors of m. Then d_(t-1) is <= n/3 and so any sum of 3 divisors of at most d_(t-1) is at most n and so that sum is counted per A093035. A sum of 3 divisors of m where one of the divisors is d_t = m as more than m so not counted. This gives (t-1)^3 possible triples hence all terms are cubes.
FORMULA
a(n) = A093035(A147516(n)).
a(n) = A000005(A147516(n) - 1)^3.
EXAMPLE
A147516(6) = 45 so a(6) = A093035(45) = (tau(45) - 1)^3.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David A. Corneth, May 19 2020
STATUS
approved